In the circuit below there is basically a resistance bridge with the inputs of an Op Amp connected across it. The output of the op amp is used to drive Q1 and would seem to create positive feed back via R5 and R2.

I suspect that this is designed to achieve constant current (or power?) through the bridge circuit but I am a bit confused as my limited knowledge of op amps suggests that the positive feed back will just cause the op amp to swing rail to rail and so just switch the transistor on or off.

Please can someone explain what Q1 is doing and how it would be affected by changing the bridge circuit resistor/thermistor values.

Thanks

 

OK, I think I have this worked out now, but curious no one replied...

 

OK, I think I have this worked out now, but curious no one replied...

Patience, grasshopper. Not all of us sit on this website all day.

 

OK, I think I have this worked out now, but curious no one replied...

hi,
Setting the ZERO pot will give Vout of ~5V.

The 50K therm is exposed to the ambient temperature.

The 2K therm is in the 'airflow' and is set for a self heating current.

With no airflow the bridge will be in balance so Vout ~ 5V.

If the 2K therm is now placed in a flowing, cooling airflow, its resistance will increase and the bridge will not be balanced so the Vout will increase.

The cooling effect is roughly proportional the rate of the air flow, so the Vout gives an an approx log indication in metres per second of airflow. The CAL pot is used to give a rough calibrated range.

The BC107 is an attempt to linearise the response of the bridge over its working range, by using feedback from Vout.

OK.?
Eric

 

Thanks Eric. The point of my question though was why isn't it unstable since there would seem to be positive feed back via R5 and R2.

I think the answer is that the transistor damps out the positive feed back, effectively creating negative feedback. If for example the output of the op amp starts to rise the transistor conducts more lowering the input voltage.

So presumably as long as the output of the op amp is somewhere in the transistors active region then it will work.

 

hi sirch,
Just tried it in LTSpice simulation, VERY unstable, the ZERO pot is so sensitive.!

By very fine tuning of the ZERO pot its possible to make the Vout ~6V, but Vout either jumps from about 0V to +9V.!

Even when when balanced, at switch ON, the Vout is a damped oscillation for many seconds.

I wondered why the designer had not added any values for the Zero and Cal pot circuits, I guess I now know why.

I will keep the LTS sim on file in case you want more tests done.

Eric

 

I think the answer is that the transistor damps out the positive feed back, effectively creating negative feedback. If for example the output of the op amp starts to rise the transistor conducts more lowering the input voltage.

I don`t think that the transistor will produce a phase inversion.
However, don`t overlook the negative feedback path.
For dc stability the negative feedback must be dominant - but this depends on the actual settings of R3 and R4.

 

Thanks for simulating that Eric, so it may be that the positive feedback is making it unstable? I wonder if changing the transistor characteristics or op amp gain would help.

I am considering using a 100R thermistor for the heated element so obviously that would change things quite a lot. I have only used LT Spice once or twice but I would be interested in having a go if you would be prepared to share your model?

 

hi,
If you are going into LTS I would recommend signing up for the free user group.
There are lots of models and tutorials that will be be useful.

I guess you are designing for airflow rate measurement.?

Can you post the Beta value for your 100R, 25Cref thermistor and the ambient temperature and airflow limits, I can then try out your LTS 'asc' files.

EDIT:
There is a thermistor Calc on this link which I designed sometime ago, you may find it helpful.
http://www.electro-tech-online.com/tools/ThermPlotV2.php

E
https://groups.yahoo.com/neo/groups/LTspice/files/%20Lib?prop=eupdatehttp://groups.yahoo.com/neo/groups/LTspice/files/%2520Lib%3fprop=eupdate

 

Hi Eric,

yes it's for detecting air flow, in fact the aim is to detect very slight drafts in caves i.e. ones that can be barely be felt by a person - so really in the range 0.1m/s to 1m/s. I have done tests that indicate the limit of perception, for me at least is around 0.3m/s. The ambient temperature is typically around 10 or 11°C.

I am thinking of a thermistor like this http://uk.rs-online.com/web/p/thermistors/0110453/
with a Beta of 3250 but it will be close limit of its power rating with 6V across the bridge. The small, SMD package means low thermal mass and better response times though.

Nice thermistor calculator by-the-way and thanks for the interest.


Chris

 

@Sirch2:

may I place some additional remarks to your circuit?

The opamp - with a very large dc open-loop gain - is connected to two voltage divider chains forming a temperature-sensitive resistive bridge (R1-R3 and R2-R4).
Both resistor/thermistor combinations are fed by the same voltage (point U).
Hence, for stable operation within the linear operating range of the amplifier the voltage across R3 must be only slightly larger than the dc voltage across R4 (in order to ensure that negative feedback dominates).
That is a very critical design (as mentioned already by Eric).
My recommendation: Use another differential amplifier with a much lower open-loop gain (with local feedback or any other differential amplification unit, or start a google search, keyword "bridge amplifier").

 

Thanks LvW, that circuit is taken from this page http://www.fonema.se/anemom/anemom.html and my understanding is that the aim of the circuit is to provide a constant current through the bridge circuit. This is necessary because the resistance of the thermistor obviously changes with temperature but the thermistor is also being heated by the current flowing through it so current regulation is supposed to keep it stable.

There are quite a few examples on the web of this kind of circuit however I am open to suggestions for better ways of providing current regulation to the bridge.


Just as an aside here is another example of a similar circuit. I actually have one of these devices and it is very sensitive although does drift around a bit.

http://moderndevice.com/product/wind-sensor/

 

Sirch2 - I know what a thermistor is and how it works.
Nevertheless, the dc stabilization criterion is to be met. And this seems to be very critical in your circuit because the differential voltage (input to the bridge amplifier) in your case is some tenth of µVolts only (dc output voltage divided by the dc open-loop gain of the opamp).
More than that, I think the circuit as shown in your first reference looks a bit different (because you "have taken" it from the referenced page).
Edit: Sorry, forget the last sentence. Initially, I have seen the 1st circuit only.

 

After reading the description of the circuit in the given reference (dimensioning and working principle) I am a bit more optimistic.
The thermistor R4 provides a special kind of negative feedback (although in the pos. feedback path) because R4 reduces its value for rising output voltage (due to more power consumption and higher temperature, hence less resistance).
In the paper, this is called "internal heating" .
Because of this feature the thermal stability is improved - if you can find a suitable thermistor. In any case, a very fine adjustment of the equilibrium is required.

 

hi Chris,
Post your draft design when ready and I will give it a run in LTS for you.

Eric

 

LvW - thanks for the input.

Eric - I'm going away for a few days so I probably won't be able to get anything done until next week. Watch this space.

 

OK, this is what I have come up with. Basically I have separated the constant current source from the bridge amplifier. It seems to model OK just varying the thermistor (labeled TR1) as a changing resistance. However I would appreciate any input about how this might work in practice. I intend to use a TLC272 op amp for U1/U2

 

hi
Downloaded your files, will check it out, will reply Monday.
E

 

Thanks Eric

 

hi Chris,
Your sim does not reflect correctly what is actually happening in your thermistor network.

Look at these two images.
One shows two thermistors , which are specified as 100R and 10K at 25C, this should give you an idea of the resistance range.
the other is your circuit with a 100R thermistor [ the type you posted the d/s for].

As you can see you must allow for self heating of the thermistors, which means that the actual resistances will be lower than the 25C spec.
[check the therm current]

Also if you use a fixed 10K, this will mean your Vout will be effected by the ambient temperature in which the 100R thermistor is operating!
ie: a change in cave ambient may give a reading which could be mis-read as airflow.

Use a 10K therm as a ambient temperature reference input.

Eric